MHB Evaluate the double sum of a product

  • Thread starter Thread starter lfdahl
  • Start date Start date
  • Tags Tags
    Product Sum
lfdahl
Gold Member
MHB
Messages
747
Reaction score
0
Evaluate the following double sum of a product:

$$\sum_{j=1}^{\infty}\sum_{n=1}^{\infty}\left(n\prod_{i=0}^{n}\frac{1}{j+i}\right)$$
 
Mathematics news on Phys.org
Hint:

The answer is: $e-1.$
 
Suggested solution:

Let
\[\alpha_j(n) = \frac{1}{j(j+1)(j+2)...(j+n)}\]
and let
\[\beta_j(n) = \frac{1}{j(j+1)(j+2)...(j+n-1)}\]

Consider the difference:

\[\beta_j(n)-\beta_{j+1}(n) = \frac{1}{j(j+1)(j+2)...(j+n-1)}-\frac{1}{(j+1)(j+2)(j+3)...(j+n)} \\\\ = \frac{j+n-j}{j(j+1)(j+2)...(j+n)} \\\\ = n \alpha_j(n)\]

Now
\[\sum_{j=1}^{\infty} \alpha_j(n) = \frac{1}{n}\sum_{j=1}^{\infty}\left ( \beta _j(n)-\beta _{j+1}(n) \right )\]

is a telescoping sum, and we get (the limit of $\beta$ is zero):

\[\sum_{j=1}^{\infty} \alpha_j(n) = \frac{1}{n}\left ( \beta _1(n)-\lim_{j \to \infty}\beta _j(n) \right ) =\frac{\beta _1(n)}{n}= \frac{1}{n \cdot n!}\]

Finally, we´re able to evaluate the given double sum above:

\[\sum_{j=1}^{\infty}\sum_{n=1}^{\infty}\left (n\prod_{i=0}^{n}\frac{1}{j+i} \right ) = \sum_{n=1}^{\infty}n \sum_{j=1}^{\infty} \alpha _j(n) = \sum_{n=1}^{\infty}\frac{1}{n!} = e-1.\]
 
Suppose ,instead of the usual x,y coordinate system with an I basis vector along the x -axis and a corresponding j basis vector along the y-axis we instead have a different pair of basis vectors ,call them e and f along their respective axes. I have seen that this is an important subject in maths My question is what physical applications does such a model apply to? I am asking here because I have devoted quite a lot of time in the past to understanding convectors and the dual...
Fermat's Last Theorem has long been one of the most famous mathematical problems, and is now one of the most famous theorems. It simply states that the equation $$ a^n+b^n=c^n $$ has no solutions with positive integers if ##n>2.## It was named after Pierre de Fermat (1607-1665). The problem itself stems from the book Arithmetica by Diophantus of Alexandria. It gained popularity because Fermat noted in his copy "Cubum autem in duos cubos, aut quadratoquadratum in duos quadratoquadratos, et...
Insights auto threads is broken atm, so I'm manually creating these for new Insight articles. In Dirac’s Principles of Quantum Mechanics published in 1930 he introduced a “convenient notation” he referred to as a “delta function” which he treated as a continuum analog to the discrete Kronecker delta. The Kronecker delta is simply the indexed components of the identity operator in matrix algebra Source: https://www.physicsforums.com/insights/what-exactly-is-diracs-delta-function/ by...

Similar threads

Back
Top